Câu hỏi của Edogawa Conan

Question

giải phương trình:

$3x^4-13x^3+16x^2-13x+3=0$

Huy Thắng Nguyễn · Accepted Answer

$3x^4-13x^3+16x^2-13x+3=0$

$\Leftrightarrow3x^4-9x^3-4x^3+12x^2+4x^2-12x-x+3=0$

$\Leftrightarrow3x^3\left(x-3\right)-4x^2\left(x-3\right)+4x\left(x-3\right)-\left(x-3\right)=0$

$\Leftrightarrow\left(x-3\right)\left(3x^3-4x^2+4x-1\right)=0$

$\Leftrightarrow\left(x-3\right)\left(3x^3-x^2-3x^2+x+3x-1\right)=0$

$\Leftrightarrow\left(x-3\right)\left[x^2\left(3x-1\right)-x\left(3x-1\right)+\left(3x-1\right)\right]=0$

$\Leftrightarrow\left(x-3\right)\left(3x-1\right)\left(x^2-x+1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-3=0\3x-1=0\x^2-x+1=0\end{matrix}\right.$

mà $x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x$

$\Rightarrow\left[{}\begin{matrix}x-3=0\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\x=\dfrac{1}{3}\end{matrix}\right.$

Vậy x = 3 hoặc x = $\dfrac{1}{3}$.Câu trả lời: $3x^4-13x^3+16x^2-13x+3=0$